Answer:
Step-by-step explanation:
I cannot use the line tool for you, but I can rewrite the equations
y = -x + 4 is good enough
Two points for this graph:
x = 0 -> y = 4 gives the point (0, 4)
x = 1 -> y = 3 gives the point (1, 3)
18x + 6y = -6
6y = -18x - 6
y = -3x - 2
Two ponts for this graph:
x = 0 -> y = -2 gives the point (0, -2)
x = 1 -> y = -5 gives the point (1, -5 )
5 Yellow = 3 Blue
2 Yellow = 3/5 *2 = 6/5 Blue, 1.2 cans of Blue .
<u>Award brainliest if helped!</u>
PEMDAS
2(12)x(12)/2
24x12/2
288/2
144
The answer is <span>√x + √y = √c </span>
<span>=> 1/(2√x) + 1/(2√y) dy/dx = 0 </span>
<span>=> dy/dx = - √y/√x </span>
<span>Let (x', y') be any point on the curve </span>
<span>=> equation of the tangent at that point is </span>
<span>y - y' = - (√y'/√x') (x - x') </span>
<span>x-intercept of this tangent is obtained by plugging y = 0 </span>
<span>=> 0 - y' = - (√y'/√x') (x - x') </span>
<span>=> x = √(x'y') + x' </span>
<span>y-intercept of the tangent is obtained by plugging x = 0 </span>
<span>=> y - y' = - (√y'/√x') (0 - x') </span>
<span>=> y = y' + √(x'y') </span>
<span>Sum of the x and y intercepts </span>
<span>= √(x'y') + x' + y' + √(x'y') </span>
<span>= (√x' + √y')^2 </span>
<span>= (√c)^2 (because (x', y') is on the curve => √x' + √y' = √c) </span>
<span>= c. hope this helps :D</span>
